The interconversion of dihydroxyacetone phosphate DHAP and g

The interconversion of dihydroxyacetone phosphate (DHAP) and glyceraldehyde - 3 - phosphate (G3P) is a part of both the glycolytic pathway and the Calvin cycle for photosynthetic carbon fixation: dihydroxyac etonephosphate ? glyceraldehyde ? 3 ? phosphate The value of ?G ° ? = +1.8 kcal/mol for this reaction at 25 ° C. In the glycolytic pathway, this reaction goes to the right, converting DHAP to G3P. In the Calvin cycle, this reaction proceeds to the left, converting G3P to DHAP.

a) In which direction does the equilibrium lie?

b) What is the equilibrium constant at 25 ° C? (Express your answer using three decimal places.)

c) In the glycolytic pathway, this reaction is driven to the right because G3P is consumed by the next reaction in the sequence, thereby maintai ning a low G3P concentration. What will ?G? be (at 25 ° C) if the concentration of G3P is maintained at 1% of the DHAP concentratio n (i.e., if [G3P]/[DHAP]=0.01)? (Express your answer using two decimal places.)

d) In the Calvin cycle, this reaction proceeds to the left. How high must the [G3P]/[DHA P] ratio be to ensure that the reaction is exergonic by at least ?3.0 kcal/mol (at 25 ° C)? (Express your answer using two decimal places.)

Solution

Answer:

a) The equilibrium lies far to the left. Because,

The ?G°’ for this reaction is positive so the Keq value is less than one.

?G°’= - (1.987 cal/mol-K) (298 K) ln Keq = +1800 cal/mol

1800/-592 = -3.04

So, Keq= 0.048.

Therefore, this equilibrium lies far to the left.

b) We can calculate the equilibrium constant K_equsing the standard free energy ?G by using the following formula.

?G = -RT In eq

Where, R is the ideal gas constant and T is temperature.

RT = -0.592 kcal mol^-1

1.8 kcal mol^-1 = (0.592 kcal mol^-1) In K_eq

K_eq = e-^3.04

K_eq = 0.048

The formula gives the equilibrium constant under standard conditions is 0.048.

(c) ?G’= ?G°’+ RT ln ([G3P] / [DHAP])

= +1800 + (1.987 cal/mol-K) (298) ln (0.01)

= +1800 + 592 (-4.605)

= 1800 – 2726

= -926 cal/mol

= -0.93 kcal/mol

(d) ?G’= ?G°’+ RT ln ([G3P] / [DHAP]). We need to find [G3P] / [DHAP] when ?G’ = -3 kcal/mol = -3000 cal/mol

-3000 = +1800 + (1.987) cal/mol-K) (298) ln ([G3P] / [DHAP])

-3000 = 1800 + 2.303 * (1.987)* (298) * log ([G3P] / [DHAP])

-3000 = 1800 + 1363.67 log [G3P] / [DHAP]

-4800 = 1363.67 log [G3P] / [DHAP]

So, log [G3P] / [DHAP] = -4800/1363.67 = -3.52

[G3P] / [DHAP] = log ^-1 (-3.52) = 0.0003